(I)设公比为q（q不=0)
2^10*S30-(2^10+1）*S20+S10
=2^10*(S30-S20)+S10-S20
=2^10*q^10(S20-S10)-(S20-S10)
=0
又因为S20-S10不=0
所以2^10*q^10-1=0
(2q)^10=1
q=0（舍去）或1/2或-1/2（因为an为正,舍去）
q=1/2
an=a1*q^(n-1)=(1/2)^n
(II)Sn=a1*[1-q^(n-1)]/(1-q)
=1-(1/2)^(n-1)
nSn=n-n(1/2)^(n-1)
设其前n项和为Tn
Tn=1S1+2S2+3S3+…+nSn
=1-(1/2)^0+2-2*(1/2)^1+3-3*(1/2)^2+…+n-n(1/2)^(n-1)
=(1+2+3+…+n)-(1/2)^0-2*(1/2)^1-…-n*(1/2)^(n-1) (1)
(1/2)Tn=(1/2)*(1+2+3+…+n)-(1/2)^1-2(1/2)^2-(n-1)*(1/2)^(n-1)-n*(1/2)^n (2)
(1)-(2)得
(1/2)Tn=(1/2)*(1+2+3+…+n)-(1/2)^0-(1/2)^1-(1/2)^2-…-(1/2)^(n-1)+n*(1/2)^n
=(1/2)*[n+n*(n-1)/2]-{1*[1-(1/2)^n]/[1-(1/2)]}+n*(1/2)^n
所以Tn=n+(n^2-n)/2-4+4*(1/2)^n+2n*(1/2)^n
=n/2+n^2/2+(2n-4)*(1/2)^n-4